Matakuliah : D0762 – Ekonomi TeknikTahun : 2009

Annual Worth AnalysisCourse Outline 5

Outline

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• Principle and Benefit• Equivalent Annual Worth • Capital Ownership Cost• AW by salvage sinking-fund method• AW by Salvage present-worth method• AW by Capital recovery plus interest method• Spreadsheet

Refererences- Engineering Economy – Leland T. Blank, Anthoy J. Tarquin

p.180-199- Engineering Economic Analysis, Donald G. Newman, p. 141-163- Engineering Economy, William G. Sulivan, p.137-194, p. 212-

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Annual Worth Analysis• Principle: Measure investment worth on annual

basis• Benefit:  By knowing annual equivalent worth, we

can:• Seek consistency of report format• Determine unit cost (or unit profit)• Facilitate unequal project life comparison

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AE(12%) = $189.43(A/P, 12%, 6) = $46.07

PW(12%) = $189.43

Computing Equivalent Annual Worth

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$100$50

$80$120

$700

2 3 4 5 61

A = $46.07

2 3 4 5 61

$189.43

0

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Annual Equivalent Worth • Repeating Cash Flow Cycles

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$500$700

$800

$400 $400 $500

$700$800

$400 $400

$1,000 $1,000

Repeating cycle

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Annual Equivalent Worth • First Cycle:

PW(10%) = -$1,000 + $500 (P/F, 10%, 1)       + . . . + $400 (P/F, 10%, 5)     = $1,155.68AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87

• Both Cycles:PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)       + . . . + $400 (P/F, 10%, 5)     = $1,873.27AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87

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Annual Equivalent Cost• When only costs are

involved, the AEmethod is called theannual equivalent cost.

• Revenues must covertwo kinds of costs:Operating costs andcapital costs.

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Capital costs

Operating costs

+

Ann

ual W

orth

Cos

ts

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Capital (Ownership) Costs• Definition: The cost of owning

an equipment is associatedwith two transactions—(1)its initial cost (I) and (2) its salvage value (S).

Capital costs: Taking intothese sums, we calculatethe capital costs as:

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CR i I A P i N S A F i NI S A P i N iS

( ) ( / , , ) ( / , , )( )( / , , )

0 1 2 3 N

0N

I

S

CR(i)

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Example - Capital Cost Calculation

• Given:       I = $200,000      N =  5 years      S = $50,000       i = 20%

• Find: CR(20%)

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157.60$

000.50$20,05%,20,/000.50$000.200$%20

,,/

PACRiSNiPASIiCR

$ 200,000

$50,000

5

0

Justifying an investment based on AE Method

• Given: I = $20,000, S =$4,000, N = 5 years, i =10%

• Find: see if an annualrevenue of $4,400 isenough to cover thecapital costs.

• Solution:CR(10%) = $4,620.76

• Conclusion: Need anadditional annual revenuein the amount of $220.76.

10http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm

Salvage -Sinking Fund Method

• General Equation

• Example 6.1Calculate the AW of a tractor attachment that has an initial cost of $8000 and a salvage value of $500 after 8 years. Annual operating cost for the machine are estimated to be $900 and an interest rate of 20% per year is applicable.Solution The problem indicates there are 2 cashflowAW = A1 + A2Where A1 = annual cost of initial investment with salvage value considered Equation above = -8000(A/P,20%8) + 500(A/F,20%,8) = $2055 A2 = annual operating cist = $-900

The Annual worth for the attachment is :AW = -2055 – 900 = $-2955 11

niFASniPAPAW ,,/,,/

Salvage Present-Worth Method

• General Equation

• The Steps to determine the complete asset AW are1. Calculate the present worth of the salvage value via the P/F

factor2. Combine the value obtained in step 1 with the investment cost

P3. Annualize the resulting difference over the life of the asset

using the A/P factor4. Combine any uniform annual worth with the value from step 35. Convert any other cash flows into an equivalent uniform annual

worth and combine with the value obtained in step 4

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niPAniFPSPAW ,,/],,/[

Example 6.2

• Compute the AW of the attachment detailed in Example 6.1 using the salvage present worth method Solution Using the steps outline and equation before AW = [-8000+500(P/F,20%,8)(A/P,20%,8) – 900

= $-2955

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Capital Recovery Plus Interest Method

• General equation

• The steps to be followed for this method are :1. Reduce the initial cost by the amount of the salvage value2. Annualize the value in step 1 using A/P factor3. Multiply the salvage value by the interest rate4. Combine the values obtained in steps 2 and 35. Combine any uniform annual amounts6. Convert all other cash flows into equivalent uniform amount and

combine them with the value from step 5Step 1 through 4 are accomplished bya pplying equation before

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iSniPASPAW ,,/)(

Example 6.1• Use the value of Example 6.1 to compute the AW

using the capital recovery plus interest

SolutionFrom equation and steps before :AW =-(8000-500)(A/P,20%,8)-500(0,20) – 900

= $-2955

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Comparing Alternatives• Following costs are estimated for two equal service tomato

peeling machines to evaluated by a canning plant manager’if the minimum required rate of return is 15% per year, help the manager decide which machine to select !

• Solution AW A = -26,000(A/P,15%,6) + 2000(A/F,15%,6)-11800 = $-18,442AW B = -36,000(A/P,15%,10)+3000(A/F,15%,10) = $-16,925

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Machine A Machine B

First Cost 26,000 36,000Annual maintenance cost,$ 800 300Annual Labor cost, $ 11,000 7000Extra annual income taxes, $ - 2,600Salvage Value 2,000 3,000Life, years 6 10

Spreadsheet

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Example 6.10If Ms.Kaw Deposits $10,000 now at an interest rate of 7% per year, hom many year s must the money accumulate before she can withdraw $1400 per year forever